Graduate Is this operator bounded or unbounded?

[SeM]

Hi, I have an operator which does not obey the following condition for boundedness:

\begin{equation*}
||H\ x|| \leqslant c||x||\ \ \ \ \ \ \ \ c \in \mathscr{D}
\end{equation*}

where c is a real number in the Domain D of the operator H.

However, this operator is also not really unbounded, because if H was unbounded, it would give an incorrect inequality, examplewise: if an bounded operator gives a valid inequality (i.e x < cx), a unbounded operator gives an invalid inequality (i.e x > cx). This operator however gives a result with complex values!

What can one say about its bounded/unbounded state then?

Thanks!

 

[George Jones]

Hi, I have an operator which does not obey the following condition for boundedness:

\begin{equation*}
||H\ x|| \leqslant c||x||\ \ \ \ \ \ \ \ c \in \mathscr{D}
\end{equation*}

where c is a real number in the Domain D of the operator H.

$c$ is not in the domain of $H$, as $c$ is just a number. $x$ is in the domain of $H$.

What does "domain" mean?

However, this operator is also not really unbounded, because if H was unbounded, it would give an incorrect inequality, examplewise: if an bounded operator gives a valid inequality (i.e x < cx), a unbounded operator gives an invalid inequality (i.e x > cx). This operator however gives a result with complex values!

I have no idea what you are trying to say. Are you saying that $c$ is complex number that is not real. If so, how can you even talk about inequalities? The set of complex numbers does not possesses a total order. Also, if $x$ is an element of a Hilbert space, the Hilbert space probably does possesses a total order. $x > cx$ has no meaning.

Note the the definition of boundedness of operators involves norms of elements of Hilbert spaces, and the norm $||x||$of Hilbert space element $x$ is a real number.

What is your particular operator?

 

[SeM]

$c$ is not in the domain of $H$, as $c$ is just a number. $x$ is in the domain of $H$.

What does "domain" mean?

Hi George, its the Kreyszig text I am looking at. p 524. He writes it precisely as so "the domain D(T) lies in a complex Hilbert space."

I have no idea what you are trying to say. Are you saying that $c$ is complex number that is not real. If so, how can you even talk about inequalities? The set of complex numbers does not possesses a total order. Also, if $x$ is an element of a Hilbert space, the Hilbert space probably does possesses a total order. $x > cx$ has no meaning.

Note the the definition of boundedness of operators involves norms of elements of Hilbert spaces, and the norm $||x||$of Hilbert space element $x$ is a real number.

What is your particular operator?

Yes, c is a complex number, and therefore I am wondering what to do with this operator in terms of using the definition of boundedness given by Kreyszig on p. 524?

The operator B = (h/i d/dx - g) , where g is a constant. If you try to solve the norm:

\begin{equation}
||Bx|| =\big( (h/i dx/dx - gx)^2\big)^{(1/2)}
\end{equation}

you get

\begin{equation}
||Bx|| = (h/i - gx)
\end{equation}and when setting up the inequality as given in definition given in the original post, you get:\begin{equation}
||Bx|| x \geqslant - ic/\gamma\hbar x
\end{equation}

I see what you mean about that this does not make any sense. So how does one prove if an operator as B is bounded or not, if the condition given in the original posting is not valid?

 

[SeM]

Are you saying that $c$ is complex number that is not real. If so, how can you even talk about inequalities?

Indeed, this is a bigger problem than I thought. Can one evaluate only the Real part in an inequality? i.e. When doing the same thing with this operator with the Schwarz inequality, the same thing happens there, complex numbers appear in the norms. How can one then determine if such an operator (B):1. Is bounded or not.
2. Follows the Schwarz inequality for the position operator or not.

?

Thanks!

 

[martinbn]

Hi, I have an operator which does not obey the following condition for boundedness:

\begin{equation*}
||H\ x|| \leqslant c||x||\ \ \ \ \ \ \ \ c \in \mathscr{D}
\end{equation*}

where c is a real number in the Domain D of the operator H.

Well, that is not what the text say. It says $c$ is a real number and that the inequality holds for all $x\in \mathscr{D}(H)$.

 

[martinbn]

The operator B = (h/i d/dx - g) , where g is a constant. If you try to solve the norm:

\begin{equation}
||Bx|| =\big( (h/i dx/dx - gx)^2\big)^{(1/2)}
\end{equation}

you get

\begin{equation}
||Bx|| = (h/i - gx)
\end{equation}and when setting up the inequality as given in definition given in the original post, you get:\begin{equation}
||Bx|| x \geqslant - ic/\gamma\hbar x
\end{equation}

I see what you mean about that this does not make any sense. So how does one prove if an operator as B is bounded or not, if the condition given in the original posting is not valid?

Here you are confusing yourself with the notations. The $x$ stands for the variable of the functions, which are the vectors in your Hilbert space, and the vector itself. Instead of $||Bx||$ write $||Bf(x)||$.

 

[SeM]

Here you are confusing yourself with the notations. The $x$ stands for the variable of the functions, which are the vectors in your Hilbert space, and the vector itself. Instead of $||Bx||$ write $||Bf(x)||$.

That is fine, I treated it as a variable when I was trying to solve the norm ||Bx|| , however there is a typo above: (6) is really:

\begin{equation}
x \geqslant - ic/\gamma\hbar x
\end{equation}, which as was said by George Jones in another thread, makes no sense. So I am not able to calculate the norm of the operator B on the variable x, and can therefore not use any of the relations (Schwarz inequality or the condition for boundedness given in the original post above) on this operator B = (ihd/dx -g).

Is there a way to determine if B is bounded/unbounded and or dependent on the position and momentum operators?

Thanks!

 

[George Jones]

The operator B = (h/i d/dx - g) , where g is a constant. If you try to solve the norm:

\begin{equation}
||Bx|| =\big( (h/i dx/dx - gx)^2\big)^{(1/2)}
\end{equation}

you get

\begin{equation}
||Bx|| = (h/i - gx)
\end{equation}

This is not correct.

which as was said by George Jones in another thread, makes no sense.

I didn't say this in another thread, I said it in this thread.

 

[SeM]

This is not correct.
I didn't say this in another thread, I said it in this thread.

Thanks, haha, sorry, many threads today.

Can you give me comment on how to determine the boundedness of this operator? George, I am looking for a good book on operator mathematics, and given that I like Kreyszig and Bohm (physics), is there a book written on that topic which is similar to this authors text in "well-readness"?

 

[George Jones]

@SeM , it is somewhat hard to tell, but I do not think that you have addressed @martinbn 's concern.

What does "domain" mean? Not just in Kreyszig, but as a general concept.

 

[SeM]

@SeM , it is somewhat hard to tell, but I do not think that you have addressed @martinbn 's concern.

This is true, and I have not found out how to solve the norm of an operator, because I considered the x as a variable, but apparently, it's rather the x coordinate in a vector set, if I am not mistaking.

What does "domain" mean? Not just in Kreyszig, but as a general concept.

The group of the entire set of elements in a given space.

 

[George Jones]

The group of the entire set of elements in a given space.

Think back to high school. What did "domain" mean in high school maths?

Also be careful with your terms. By "group", I think that you mean "collection". "group" means something specific in abstract algebra.

 

[SeM]

Think back to high school. What did "domain" mean in high school maths?

Also be careful with your terms. By "group", I think that you mean "collection". "group" means something specific in abstract algebra.

Sorry, I intended a set, or a collection, but also a set is something, so your word collection is the best.

I didnt learn domain in High school but going back to the first time I heard it, I recalll a geometrical space that was confined by a fence , or better, a boundary.

 

[George Jones]

I didnt learn domain in High school

Did you see notation like
$$f : A \rightarrow B$$
in high school or early university? If so, what does this mean?

 

[SeM]

Did you see notation like
$$f : A \rightarrow B$$
in high school or early university? If so, what does this mean?

No, I haven't. From what I have learned SOLELY on physicsforum, I take that it means that some operator f is transforming a function from a space A to a space B.

 

[George Jones]

Did you see notation like
$$f : A \rightarrow B$$
in high school or early university? If so, what does this mean?

No, I haven't.

If you have not seen this, I have to ask: what mathematics courses have you taken in school?

 

[SeM]

If you have not seen this, I have to ask: what mathematics courses have you taken in school?

Norrway is a country that hates Mathematics, that is the reason.

I have taken engineering mathematics at graduate level, 2 years, then I started an Advanced Mathematics course during my PhD 2 years ago, which I enjoyed a lot and got 5 out of 6 on, by working from home and taking the exam at the Uppsala University in sweden, where I take my PhD nowadays. All this interest in math is for own stuff.